Lecture Note on Linear Algebra 15. Dimension and Rank


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1 Lecture Note on Linear Algebra 15. Dimension and Rank WeiShi Zheng, 211 November 1, What Do You Learn from This Note We still observe the unit vectors we have introduced in Chapter 1: 1 e 1 =, e 2 = 1, e 3 =. (1) 1 We know the above are the basis (specially the standard basis) of R 3. However, we still have to answer the following question: Question: Why there are three basis in R 3? Basic Concept:dimension( 维 数 ), rank( 秩 ) 1
2 2 Dimension Theorem 1. Let V be a vector space with basis B = { b 1,..., b n }. Then any subset of V containing more than n vectors is linearly dependent. Proof. Let p N and p > n. Assume that { u 1,..., u p } V. As B is the basis, so we have for each u i there exists a coefficient vector a i R n such that u i = B a i, B = [ b 1,..., b n ]. Let A = [ a 1,..., a p ], U = [ u 1,..., u p ]. Then U = BA. The if there is a series of weight c 1,, c p such that c 1 u c p u p =, that is U c =, then we have BA c =. As the number of rows is smaller than the number of columns of BA, so that is there are nonpivot columns in BA. This leads to the matrix equation BA c has nontrivial solution. So there exists nonzeros c such that U c =. That is u 1,..., u p are linearly dependent. Theorem 2. Let V be a vector space with bases B, C of sizes m, n N respectively. Then m = n. That is every basis of V has the same size. Proof. Since C is linearly independent, by Theorem 1, we must have m n. Similarly, n m. So m = n. Definition 3 (dimension). Let V be a vector space with a finite subset as its basis. Then the size of its basis is called the dimension of V and is denoted by dim V, and we say V is finitedimensional. Otherwise, if V can not be spanned by a finite set then V is infinitedimensional. Remark: (1) dim R n = n. (2) The space with all polynomials ( 多 项 式 空 间 ) is infinitedimensional. 2
3 3 Subspace & Dimension Theorem 4. Let V be a vector space of dimension n N. Then any linearly independent subset { v 1,..., v m } of V can be extended to a basis of V. Proof. We prove this theorem by the following steps: 1. When m = n, because if Span{ v 1,..., v m } = V then { v 1,..., v m } is a basis of V. 2. When m < n Span{ v 1,..., v m } V and we can pick v m+1 V Span{ v 1,..., v m }. We claim that { v 1,..., v m, v m+1 } is linearly independent. Assume this is not the case, then there exist c 1,..., c m, c m+1 which are not all zero such that c 1 v c m v m + c m+1 v m+1 =. If c m+1 = then { v 1,..., v m } is linearly dependent, otherwise v m+1 Span{ v 1,..., v m }. Both cases contradict the assumptions we made. So { v 1,..., v m, v m+1 } is linearly independent. 3. If { v 1,..., v m, v m+1 } spans V then we obtain a basis, otherwise repeat this process again and finally, we must obtain a linearly independent set containing { v 1,..., v m } spans V, that is a basis. Theorem 5. Let H be a subspace of a finitedimension vector space V. Any linearly independent set in H can be expanded, if necessary, to a basis for H. Also, H is finitedimensional and dimh dimv Theorem 6. Let V be a vector space of dimension p( 1). Any linearly independent set of exactly p elements in V is automatically a basis for V. Any set of exactly p elements that spans V is automatically a basis for V. Proof. By theorem 4, any linearly independent set S of size q( p) can be expanded to the basis of V. This implies, if q = p, then S must be the basis of V. Theorem 7. The dimension of NulA is the number of free variables in the equation A x =, and the dimension of ColA is the number of pivot columns in A. 3
4 4 Rank of a Matrix A 4.1 Row Space Definition 8 (Row Space). If A is an m n matrix. The set of all linear combinations of the row vectors is called the row space of A, denoted by RowA. From another point of view, the rows of A are identical to the columns of A T, so the row space of A can also be written as ColA T. Theorem 9. If two m n matrices A and B are row equivalent, their row spaces are the same. Proof. Since A and B are row equivalent, there exists a invertible matrix G such that B = GA. That is B T = A T G T = A T F, F = G T. By using the matrix partition theory, let A T = [Col 1 A T,, Col m A T ], F = [Col 1 F,, Col m F ], then That is [Col 1 B T,, Col i B T,, Col m A T ] =B T =A T F =[A T Col 1 F, A T Col i F,, A T Col m F ]. Col i B T =A T Col i F =[Col 1 A T,, Col m A T ]([F ] 1i,, [F ] mi ) T =[F ] 1i Col 1 A T + + [F ] mi Col m A T. Since (Row i A) T = Col i A T and (Row i B) T = Col i B T, so Row i B = [F ] 1i Row 1 A + + [F ] mi Row m A. That is RowB RowA. Conversely, by changing the roles of A and B in the above, we can also have RowB RowA. So, RowB = RowA. Theorem 1. If two m n matrices A and B are row equivalent, and B is in echelon form, then the nonzero rows of B form a basis for the row space of A as well as for that of B (2) (3) 4
5 4.2 Definition and Properties of Rank Definition 11 (column & row rank). Let A be a m n matrix. The column rank of A is defined to be dim Col(A) and the row rank of A to be dim Row(A). Remark: The dimension of the null space is sometimes called the nullity of A. Theorem 12 (Rank Theorem). The dimensions of the column space and the row space of an m n matrix A are equal. Also, it holds that: ranka + dim NulA = n. (4) Proof. We know (1) ranka is the number of pivot columns in A. That is if B is an echelon form of A. i.e. ranka is the number of pivot positions in B. (2) Each pivot position corresponds to a nonzero row, and these rows form a basis for the row space of A, so the rank of A is also the dimension of the row space. (3) The dimension of NulA equals the number of free variables in the equation A x =. That is the dimension of NulA is the number of columns of A that are not pivot columns. (4) Number of pivot columns + number of nonpivot columns = number of columns, i.e. ranka + dim NulA = n. 5
6 5 Computation of Null Space and Rang S pace We are now discussing how to compute bases of Col(A) and Nul(A) given A R m n. Computation of Column Space: Let A = ( a 1 a n ) R m n. Then there exists an invertible matrix B R m m such that BA is in REF. Let B a i1,..., B a ir be all the pivot columns of BA. Then it is obvious that {B a i1,..., B a ir } is a basis of Col(BA). Since B is invertible and by Theorem 15.4, { a i1,..., a ir } forms a basis of Col(A), that is the set of pivot columns of A is a basis of Col(A). Computation of Null Space: We also know that Nul(A) is exactly the solution set of the equation A x =. So by solving Ax = using Row Reduction Algorithm, we obtain v 1,..., v n r Nul(A), such that Nul(A) = {x j1 v x jn r v n r x j1,..., x jn r R} = Span{ v 1,..., v n r }, where x j1,..., x jn r correspond to the free variables of Ax =. But by the Rank Theorem, dim Nul(A) = n dim Col(A) = n r. Thus { v 1,..., v n r } is a basis of Nul(A) by Theorem 5. Examples: Textbook P.24, P.241, P
7 6 Rank & Matrix Inverse Theorem 13 (The Invertible Matrix Theorem). Let A be an n n matrix. The following statements are each equivalent to the statement that A is an invertible matrix: (1) The columns of A form a basis of R n ; (2) ColA = R n ; (3) dim ColA = R n ; (4) ranka = n; (5) NulA = { }; (6) dim NulA = ; Flora, by Titian 7
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